Is ${636474}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {636474}= &&{6}\cdot100000+ \\&&{3}\cdot10000+ \\&&{6}\cdot1000+ \\&&{4}\cdot100+ \\&&{7}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {636474}= &&{6}(99999+1)+ \\&&{3}(9999+1)+ \\&&{6}(999+1)+ \\&&{4}(99+1)+ \\&&{7}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {636474}= &&\gray{6\cdot99999}+ \\&&\gray{3\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{4\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {6}+{3}+{6}+{4}+{7}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${636474}$ is divisible by $9$ if ${ 6}+{3}+{6}+{4}+{7}+{4}$ is divisible by $9$ Add the digits of ${636474}$ $ {6}+{3}+{6}+{4}+{7}+{4} = {30} $ If ${30}$ is divisible by $9$ , then ${636474}$ must also be divisible by $9$ ${30}$ is not divisible by $9$, therefore ${636474}$ must not be divisible by $9$.